What are three consecutive integers whose product is 480 more than their sum?

The three numbers are 7 8 and 9

Step-by-step explanation:

Givens

Equation

(n - 1)(n)(n + 1) - (n-1 + n + n+1) = 480

Solution

Multiply (n - 1) and (n + 1) = (n - 1)*(n + 1) = n^2 - 1

Multiply the second integer by the result of the first and third: n (n^2 - 1)

Add the three integers together: (x - 1) + (n - 1) + n = 3n  Combine these 2 steps

n(n^2 - 1) - 3n = 480 Remove the brackets

n^3 - n - 3n = 480

n^3 - 4n = 480

n^3 - 4n - 480 = 0

Graph

The graph shows that the intercept point is n =8. This is the only way I can see to solve this cubic. There are no other real roots.

Answer

n - 1 = 7

n = 8

n + 1 = 9

Check

Product 7*8*9 = 504

Sum = 7 + 8 + 9 = 24

504 - 24 = 480 Which checks.

5/12, or 5: 12

step-by-step explanation:

i think is g(1)=-1.

hope that i can